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In summary: I'm not sure how you would solve for that equation.There's also a term proportional to 1/(x-2). This gives a 2nd equation to use to solve for a,b. I'm not sure how you would solve for that equation. In summary, Homework Statement Determine all the numbers a and b for which the limit of the function as x approaches 2 equals to 4 equals 4.

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- #2

fzero

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Does writing

[tex]\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3[/tex]

give you any ideas?

[tex]\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3[/tex]

give you any ideas?

- #3

eibon

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try multiply by its Conjugate that will help you

- #4

3ephemeralwnd

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Ive gotten to a point where (x-2) cancels out, and i can sub in x=2 without the denominator coming out as undefined..

But I'm still not sure where to go from here

- #5

fzero

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3ephemeralwnd said:

Ive gotten to a point where (x-2) cancels out, and i can sub in x=2 without the denominator coming out as undefined..

But I'm still not sure where to go from here

Can you explain a bit more of your work? If you're using my hint then you expanded the square root in powers of (x-2) and there's a divergent term, a term that's independent of x, and then terms that vanish as x -> 2. The coefficients of the first 2 terms can be thought of as a pair of equations to solve for a and b once you demand that the divergent part vanishes and that the finite term gives the correct limit.

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nddancer

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fzero said:Does writing

[tex]\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3[/tex]

give you any ideas?

Hi fzero, how did you get that second part?

Thanks!

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nddancer

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fzero

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nddancer said:

All I did was add and subtract 2 a under the square root so that I could express the numerator in terms of the x-2 that appeared in the denominator. However, if you're not familiar with series expansions, the next step would be far from obvious.

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nddancer

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- #10

fzero

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nddancer said:

There's also a term proportional to 1/(x-2). This gives a 2nd equation to use to solve for a,b.

The purpose of solving for constants a and b is to determine the specific values that will make the function approach a certain limit. This can help us understand the behavior of the function and make predictions about its values at different points.

To solve for constants a and b, we use algebraic techniques such as substitution, elimination, and manipulation of equations. We can also use graphical methods to visually determine the values.

Some common strategies for solving for constants a and b include using the definition of a limit, factoring, and setting up a system of equations. It is also helpful to use known properties of functions, such as symmetry and continuity, to simplify the problem.

Yes, calculus can be used to solve for constants a and b given the limit of a function. This involves using techniques such as differentiation and integration to manipulate the function and determine the values of a and b.

There may be limitations to solving for constants a and b when dealing with more complex functions or when the limit is undefined. In these cases, it may be necessary to use numerical methods or approximation techniques to find the values of a and b.

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